MAT102H5 Lecture Notes - Lecture 2: Quadratic Equation, Quadratic Formula

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16 Sep 2018
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MAT102H5 Full Course Notes
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Theorem 1: quadratic formula if: following holds true: (cid:1853)2+(cid:1854)+(cid:1855)=0, where (cid:1853),(cid:1854),(cid:1855) are real numbers and (cid:1853) 0. Let (cid:1853),(cid:1854),(cid:1855) be any real number such that (cid:1853) 0. Consider (cid:1853)2+(cid:1854)+(cid:1855)=0, then the: the above equation has no real roots if: (cid:1854)2 4(cid:1853)(cid:1855)<0, the above equation has a unique root given by = (cid:3029)2(cid:3028) if: (cid:1854)2 4(cid:1853)(cid:1855)=0, the above equation has two distinct roots given by = (cid:3029) (cid:3029)2 4(cid:3028)(cid:3030) Divide the equation (cid:1853)2+(cid:1854)+(cid:1855)=0 by (cid:1853) (cid:4666)(cid:1853) 0(cid:4667), then we can get. Then we set (cid:1827)=, (cid:1828)= (cid:3029)2(cid:3028), (cid:1828)2= (cid:3029)24(cid:3028)2 , then we can get: 2+(cid:1854)(cid:1853)+(cid:1855)(cid:1853)=(cid:4666)2+(cid:1854)(cid:1853)+(cid:1854)24(cid:1853)2(cid:4667) (cid:1854)24(cid:1853)2+(cid:1855)(cid:1853)=0 (since (cid:1828)2 don"t exist in the o(cid:396)iginal e(cid:395)uation, we add it and subtract it to maintain. And a step further, we can get (cid:4666)+ (cid:3029)2(cid:3028)(cid:4667)2 (cid:3029)24(cid:3028)2+(cid:3030)(cid:3028)=0 (this form is better since the variable only exists once) *proof done: if (cid:1854)2 4(cid:1853)(cid:1855)<0, then there are no real solutions since (cid:4666)+ (cid:3029)2(cid:3028)(cid:4667)2>0 but.

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